QUESTION EXPLANATIONS

1. (C) — This problem is asking you to solve for x. To do that, we just need to get x alone on one side of the equation. We can begin by rewriting the right side of the equation using the distributive property. This give us: 42 = 3x - 12. Next, we can add 12 to both sides. That gives us: 54 = 3x. Now, we just need to divide both sides by 3. That will give us 54/3 = x. 54/3 = 18, so our answer is (C).

2. (C) — This problem is asking us to find the value of a variable, k, in a quadratic equation. To accomplish this, all we need to do is rewrite the right side of the equation: (x+3)^2. We can begin by writing out that exponent: (x+3)^2 = (x+3)(x+3). Next, use FOIL to get rid of the parentheses, and group like terms: (x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x +9. And there you have it: x^2 + kx + 9 = x^2 + 6x +9, so k = 6, which is answer option (C).

3. (B) — This problem is asking you to determine the value of x in terms of y, or in other words, to find an expression with y that is equal to x. The solution to the system of equations on a graph is the point at which the lines of the two equations meet. We can see the units on the graph are given in intervals of 1, and we can pencil in the unlabeled units and determine that the point where the two lines intersect is (2, 3). But we need the value of x in terms of y, so to get there we can do a little algebra: xy = yx, so we can just take that equation and plug in the value we’ve obtained for y on one side and the value we’ve obtained for x on the other, to get 2y = 3x. To get x alone, we just divide both sides by 3 to get x = 2y/3, or 2/3y, which is (B).

4. (C) — This problem is asking you to find the profit for selling one barrel of oil. There are different costs for each step in the process, and some steps occur multiple times, so in order to solve without losing track of anything we can write an algebraic expression. The revenue for the sale, r, is the amount the barrel is sold for, or $93. The costs are: $51 for extraction, e, two transportation steps, t, at $6 each, and three processing steps, p, at $9 each. So we can represent the whole formula for profits as: p = r - e - 2t -3p. We can now plug in the values for each of our variable and solve: p = (93) - (51) - 2(6) - 3(9) = 93 - 51 - 12 - 27 = 42 - 12 - 27 = 30 - 27 = 3, which is answer option (C).

5. (D) — This problem is asking you to solve for the value of an expression including c, and in order to do that you first need to find the value of c using the equation c - 1 = 3. Just add 1 to each side: that gives us c = 4. Now we can plug 4 into c^2 - 1 to get 4^2 - 1, which equals 16 - 1, which is 15, which is answer option (D).

6. (B) — This problem asks for the area of the shaded area on the diagram. There’s a long way to solve this, and a short way. First, try drawing a line between Q and R so that you’re dealing with triangles. Next, since we have the area of the square, you could find the dimensions of the square (10x10) by taking the square root of its area (100), then note that if Q is the midpoint of A and C that means that the shaded area can be divided into two right triangles, each of which has one leg, a, of length 10 and another, b, of length 5. Since the area of a right triangle equals (ab)/2, the area of two triangles of equal size is two times (ab)/2 or just ab, which in this case is 50, which is answer option (B).

7. (A) — This problem is asking you to solve for a in an equation that also includes the variable b, and gives us the value of b. Whenever you’re given that kind of information, it’s sensible to start by plugging it in, so let’s plug it in and take another look at the equation: 2(3a - 6) = 4 x 6. Next, let’s rewrite the equation, using the distributive property on the left side: 6a - 12 = 24. Our goal is to get a alone on one side of the equation, so next add 12 to both sides: 6a = 36. Finally, divide both sides by 6: a = 6, which is answer option (A).

8. (B) — The fastest way to solve this problem is by plugging in some arbitrary value for x (but not 1, 0, or -1) and see what we get. Let’s try plugging in 3. If x = 3, then: 2x/(x - 1) - 3x/(x + 1 ) = 6/2 - 9/4 = (12 - 9)/4 = 3/4. So now, we just need to find the option that’s equal to ¾. (A), (C), and (D) are all negative, so we can rule those all right out. Plugging 3 into (B) shows that it does indeed equal ¾, so we can confirm that it’s the correct answer.

9. (C) — This problem is asking you to identify the algebraic expression that correctly expresses Joel’s age in terms of a, the current age difference between Joel and another person, Luca, and b, the number of years until Joel will be twice as old as Luca. Let’s start by creating some additional variables: we’ll use J for Joel’s age, and L for Luca’s age. Since Joel is a years older than Luca, we can write that L + a = J. Since Joel will be twice as old as Luca in b years, we can write that J + b = 2(L + b). Keep in mind that we need to have b on both sides, because both Joel and Luca are both aging at the same rate! We want to express Joel’s age, J, in terms of a and b, so we need to get rid of L. We can do that by rewriting L in terms of J: since L + a = J, L = J - a. So, let’s plug that in to get L out of our equation: J + b = 2(J - a + b). Next, rewrite using the distributive property: J + b = 2J -2a +2b. Now, we want to get J on one side of the equation. One way to do this is to subtract 2J from both sides: -J +b = -2a + 2b. To get J alone, let’s subtract b from both sides: -J = -2a + b. Finally, since J is currently negative, multiply both sides by negative 1 to get our signs straight: J = 2a - b, which is answer option (C).

10. (A) — This problem is asking you to find an inequality which represents the same range of possible values as a given inequality. The given inequality is an absolute value inequality, but none of our answer options are, so it’s reasonable to start by trying to write the absolute value function out of the equation. The value inside of the absolute value signs may be as great as 5 or as small as -5, so we can rewrite the inequality as -5 <= x-3 <= 5. Unfortunately, that’s not an answer option. However, we can rewrite this inequality to get x alone by adding three to each part of it: -2 <= x <= 8. And, there’s our answer: (A).

11. (B) — This questions is asking you to pick out a value, in terms of f (x), that is equal to f (5). This question might look a little intimidating at first, and you might be tempted to try to work out the what the function is with the information given. We know that a quadratic function will have the form ax^2 + bx + c, and given a set of points we can work out the values for that function. However, that would not be a good use of your time in this case, because there’s a much easier way to work this out. Quadratic functions are symmetrical, and that implies that points which are an equal distance from the minimum point in the positive and negative directions along the x axis will have equal values for y. In other words, if the minimum point of a quadratic equation sits at f (x), then f (x+1) = f(x-1), f (x+2) = f (x-2), etc. In this case, the minimum is at f(2). That means that, for example, f (3) = f (1). All you have to do is find a value for x that has the same difference from 2 as 5 does. 5 is 3 greater than 2, so you’re looking for a value that’s 3 less that 2, and that’s -1. So, for a quadratic equation with a minimum point at f(2), f(5) = f(-1), which is (B).

12. (B) — This question is asking you to determine the value of x in terms of a. An equation is provided, in which a and x are both exponents of terms in a fraction that’s equal to ¼. Notice that the denominator actually equals 4 × 4^a, so we can write the equation as 16^x/4(4^a) = ¼. Multiplying both sides by 4(4^a) gives us 16^x = 4^a. We can simplify the left side to 4^2x (since 16 = 4²), giving us 4^2x = 4^a. From here we can see that 2x = a, so x = a/2.

13. (D) — This problem is asking you to find the value of a, and in order to do that you have to make some inferences with the information that we’re given. There are a few ways to solve this, but a sensible starting point is to write some of the information that we’re given in an algebraic form. We know that: a + b = 132. Since we know that the product of a and b is negative, which implies that one of them is a negative number, but that a is the square of b, which implies that a is a positive number, we know that b is negative. So, b < 0 < a. Because b < 0, a + b < a, so 132 < a. Or, to put that in really plain English: once we know that b is negative, we know that a has to be greater than the sum of a and b. So, the only possible answer option is (D).

14. (A) — This problem asks us to find the value for x, where x is part of a term for the length of one side of a triangle. You can recognize that several of the given values for x are implausibly large: 10x = 50, and 50 is greater than the sum of 15 and 20. One side of a triangle cannot have a length that is greater than the sum of the other two sides, so this is not a possible value for x. Since (C) and (D) are even greater than (B), only (A) is a possible value for x.

15. (B) — The system of equations provided allows us to solve for a numerical value of x and y. Isolating y in the second equation and substituting the equations in, we get 7x – 6 = 5x^2 – 3x – 1. When we move all the terms to one side and factor the quadratic equation, we get 0 = 5(x – 1)^2, meaning x = 1. Plugging in the value of x to either of the equations, we find that y = 1. Because y and x both equal 1, y = x. The correct answer is (B).

16. (3) — This problem describes a function that models the dropping of a stone, and asks you to solve for the number of seconds that it takes for the stone to hit the ground. To solve this problem, we need to take a closer look at the function h(t) = a - t^2. We’ve been told that t = time, and that h(t) = the height of the stone, but we don’t know what a is. We’ve also been told that the stone is dropped from a height of 9 meters, or in other words, that at t = 0 the height of the stone was 9. We can write that in terms of our function as follows: h (0) = a - 0^2 = 9, which tells us that a = 9. Next, we just have to figure out for what t h (t) = 0, or in other words, solve for t in 9 - t^2 = 0. We can add t^2 to both sides to get 9 = t^2, and because 9 is a perfect square we can easily calculate that the square root root of 9 = 3 = t, which gives us our answer: 3.

17. (10) — This problem asks us to determine the length of a line segment, given expressions for the lengths of two other line segments. The first thing is to work out what the relationship of line BC is to AB and AC. BC is a part of line AC: it starts where AB ends, and ends where AC ends. That means that the length of BC will be equal to the difference between the lengths of AB and BC. We don’t need to know what x is in order to work out the difference between x - 4 and x + 6: that’s 10. Since BC is equal to the difference between AB and AC, BC = 10.

18. (17/2) — We can solve for the numerator by plugging 2 into the function f (x), giving us f (2) = 8(2) +1 = 17. To solve for the denominator, we first find f (0), which is 8(0) +1 = 1. g (f (0)) is therefore g (1) = 3(1) – 1 = 2. When we put the fraction together, we get 17/2.

19. (6) — The fastest way to solve this problem is by plugging in values. This question is asking us to find the value of d, given a pair of equations. The problem also stipulates that d is a positive number. We can solve for y using the second equation, then plug our value for y into the first equation in order to solve for d. Since 6y -9 = y^2, we know that 6y is > y^2. That implies that y is less than 6. We also know that y is positive, so there’s just a small range of possible values for y remaining, and we can make quick work of this equation by just plugging in some possible values. It’s not 1, because 6 - 9 is -3 and no real number squares to -3. It’s not 2, because 2^2 = 4 while 12 - 9 = 3. Plugging in 3, we can see that 3^2 = 9 and 18 - 9 = 9, so y = 3. We can plug this value into the first equation and d^2= 12(3) = 36, d = 6.

20. (6) — The problem is asking us to solve for the value of an expression that includes an imaginary number, which might be a little intimidating if you haven’t worked with imaginary numbers much before. However, the prompt actually gives you all the information that you need about i in order to solve the problem. Just start out doing what you would do if i were any other number, and rewrite this using FOIL: . Once you take this step, you can see that i has transformed into i^2--which the prompt has already told us is equal to -1. So now, we can just substitute -1 for i in the expression, and solve: 1 - (-1)5 = 1 + 5 = 6.